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Binomial theorem to evaluate limits?

  1. Oct 11, 2013 #1
    1. The problem statement, all variables and given/known data

    lim x->1 (X^9 + x -2)/(x^4 + x -2)

    I know how to do this using L'Hopitals Rule and I get 2

    2. Relevant equations

    (1+b)^n = 1 + bn + n(n-1)b^2/2! + n(n-1)(n-2)b^3/3! ....

    3. The attempt at a solution

    Let x = h+1

    x -> 1
    h -> 0

    lim h->0 (h+1)^9 + h-1/(h+1)^4+h-1

    lim h->0 h^9 (1+1/h)^9 +h-1/h^4(1+1/h)^4 +h-1

    Binomial theorem:

    (1+1/h)^9 = 1 +9/h +36/h^2 + 84/h^3 ...

    (1+1/h)^4 = 1 +4/h + 6/h^2 + 4/h^3 ...

    lim h->0 h^9(1 +9/h +36/h^2 + 84/h^3 ...) +h-1/h^4(1 +4/h + 6/h^2 + 4/h^3 ...) +h-1


    This is stil ∞.0 right?

    I tried to get rid of the +h-1 at the end by doing

    h^7(h^2 +10h +35 +84/h...)/h^2(h^2+5h+5+4/h...)

    but then you get ∞.∞
     
  2. jcsd
  3. Oct 11, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You are fine to
    lim h->0 ((h+1)^9 + h-1)/((h+1)^4+h-1)
    But I don't understand why you then switched to "1/h". Taking h to 0 is much simpler than taking 1/h to infinity. By the binomial theorem, (h+ 1)^9= h^9+ 9h^8+ ...+ 36h^2+ 9h+ 1 so that (h+1)^9+ h- 1= h^9+ 9h^8+ ...+ 36h^2+ 10h. Similarly (h+ 1)^4= h^4+ 4h^3+ 6h^2+ 4h+ 1 so that (h+ 1)^4+ h- 1= h^4+ 4h^3+ 6h^2+ 5h.

    The fraction can be written
    [tex]\frac{h^9+ 9h^9+ ...+ 36h^2+ 10h}{h^4+ 4h^3+ 6h^2+5h}= \frac{h(h^8+ 9h^7+ ...+ 36h+ 10)}{h(h^3+ 4h^2+ 6h+ 5)}[/tex]
    Now those first "h" terms cancel and you can just take h= 0.

    Are you required to use that method? Simpler than that or L'Hopital:

    The only problem here is that both numerator and denominator are 0 at x= 1. But since they are polynomials, that means that each has x- 1 as a factor:
    [tex]x^9+ x- 2= (x- 1)(x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2)[/tex]
    and
    [tex]x^4+ x- 2= (x- 1)(x^3+ x^2+ x+ 2)[/tex]
    So you just need to evaluate
    [tex]\frac{x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2}{x^3+ x^2+ x+ 2}[/tex]
    at x= 1.
     
  4. Oct 11, 2013 #3
    How did you divide x^4 +x -2 by (1-x)? When I did it I got (-x^3+x^2+x) -2/(1-x)
    I did the polynomial division method.

    Also, thank you.
     
  5. Oct 12, 2013 #4
    Sorry, I mean how did you divide x^4 +x -2 by (x-1)? I got (x^3 -x^2 -x -2/(x-1) )

    and for x^9 +x -2 divided by (x-1) I got (x^8-x^7 -x^6 -x^5 -x^4 -x^3 -x^2 -x -2/(x-1))
     
  6. Oct 13, 2013 #5

    epenguin

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    Homework Helper
    Gold Member

    Whether they are true is something you can more easily check than how it was got.
     
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